Mixing
Searching Shift/Permutation Matrix
Clash Royale CLAN TAG#URR8PPP
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Is there a Matrix P, which does:
$$AP=A'$$
$$A=beginbmatrix
a_00 & 0 & 0\
a_10 &a_11 & 0\
a_20 &a_21 & a_22
endbmatrix$$
$$A'=beginbmatrix
a_00 & a_11 & a_22\
a_10 & a_21& 0\
a_20 & 0 & 0
endbmatrix$$
linear-algebra permutations
edited Jul 26 at 9:30
Bill Wallis
1,7811821
asked Jul 25 at 14:49
Bongo1234
11
 |Â
show 1 more comment
up vote
0
down vote
favorite
1
Is there a Matrix P, which does:
$$AP=A'$$
$$A=beginbmatrix
a_00 & 0 & 0\
a_10 &a_11 & 0\
a_20 &a_21 & a_22
endbmatrix$$
$$A'=beginbmatrix
a_00 & a_11 & a_22\
a_10 & a_21& 0\
a_20 & 0 & 0
endbmatrix$$
linear-algebra permutations
edited Jul 26 at 9:30
Bill Wallis
1,7811821
asked Jul 25 at 14:49
Bongo1234
11
What are your thoughts on the problem?
– Parcly Taxel
Jul 25 at 14:57
1
How would you solve $AB=C$ for $B$?
– Ross Millikan
Jul 25 at 15:00
For $A$ invertible, i.e. $a_00cdot a_11cdot a_22neq 0$, i.e. for $a_00, a_11, a_22neq 0$, you can solve the equation $AP=A'$ with $A^-1AP=A^-1A'$, i.e. $P=A^-1A'$.
– zzuussee
Jul 25 at 15:07
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and $$A^Ntimes N$$?
– Bongo1234
Jul 25 at 15:24
Try to check the conditions on the matrix $P$ by evaluating the product by hand for the entries in $A'$ this should give you a linear system of equations in variables of $P$ which you then can check for solvability.
– zzuussee
Jul 25 at 18:50
 |Â
show 1 more comment
up vote
0
down vote
favorite
1
up vote
0
down vote
favorite
1
1
Is there a Matrix P, which does:
$$AP=A'$$
$$A=beginbmatrix
a_00 & 0 & 0\
a_10 &a_11 & 0\
a_20 &a_21 & a_22
endbmatrix$$
$$A'=beginbmatrix
a_00 & a_11 & a_22\
a_10 & a_21& 0\
a_20 & 0 & 0
endbmatrix$$
linear-algebra permutations
edited Jul 26 at 9:30
Bill Wallis
1,7811821
asked Jul 25 at 14:49
Bongo1234
11
Is there a Matrix P, which does:
$$AP=A'$$
$$A=beginbmatrix
a_00 & 0 & 0\
a_10 &a_11 & 0\
a_20 &a_21 & a_22
endbmatrix$$
$$A'=beginbmatrix
a_00 & a_11 & a_22\
a_10 & a_21& 0\
a_20 & 0 & 0
endbmatrix$$
linear-algebra permutations
edited Jul 26 at 9:30
Bill Wallis
1,7811821
asked Jul 25 at 14:49
Bongo1234
11
edited Jul 26 at 9:30
Bill Wallis
1,7811821
edited Jul 26 at 9:30
Bill Wallis
1,7811821
edited Jul 26 at 9:30
Bill Wallis
1,7811821
1,7811821
asked Jul 25 at 14:49
Bongo1234
11
asked Jul 25 at 14:49
Bongo1234
11
asked Jul 25 at 14:49
Bongo1234
11
11
What are your thoughts on the problem?
– Parcly Taxel
Jul 25 at 14:57
1
How would you solve $AB=C$ for $B$?
– Ross Millikan
Jul 25 at 15:00
For $A$ invertible, i.e. $a_00cdot a_11cdot a_22neq 0$, i.e. for $a_00, a_11, a_22neq 0$, you can solve the equation $AP=A'$ with $A^-1AP=A^-1A'$, i.e. $P=A^-1A'$.
– zzuussee
Jul 25 at 15:07
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and $$A^Ntimes N$$?
– Bongo1234
Jul 25 at 15:24
Try to check the conditions on the matrix $P$ by evaluating the product by hand for the entries in $A'$ this should give you a linear system of equations in variables of $P$ which you then can check for solvability.
– zzuussee
Jul 25 at 18:50
 |Â
show 1 more comment
What are your thoughts on the problem?
– Parcly Taxel
Jul 25 at 14:57
1
How would you solve $AB=C$ for $B$?
– Ross Millikan
Jul 25 at 15:00
For $A$ invertible, i.e. $a_00cdot a_11cdot a_22neq 0$, i.e. for $a_00, a_11, a_22neq 0$, you can solve the equation $AP=A'$ with $A^-1AP=A^-1A'$, i.e. $P=A^-1A'$.
– zzuussee
Jul 25 at 15:07
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and $$A^Ntimes N$$?
– Bongo1234
Jul 25 at 15:24
Try to check the conditions on the matrix $P$ by evaluating the product by hand for the entries in $A'$ this should give you a linear system of equations in variables of $P$ which you then can check for solvability.
– zzuussee
Jul 25 at 18:50
What are your thoughts on the problem?
– Parcly Taxel
Jul 25 at 14:57
What are your thoughts on the problem?
– Parcly Taxel
Jul 25 at 14:57
1
1
How would you solve $AB=C$ for $B$?
– Ross Millikan
Jul 25 at 15:00
How would you solve $AB=C$ for $B$?
– Ross Millikan
Jul 25 at 15:00
For $A$ invertible, i.e. $a_00cdot a_11cdot a_22neq 0$, i.e. for $a_00, a_11, a_22neq 0$, you can solve the equation $AP=A'$ with $A^-1AP=A^-1A'$, i.e. $P=A^-1A'$.
– zzuussee
Jul 25 at 15:07
For $A$ invertible, i.e. $a_00cdot a_11cdot a_22neq 0$, i.e. for $a_00, a_11, a_22neq 0$, you can solve the equation $AP=A'$ with $A^-1AP=A^-1A'$, i.e. $P=A^-1A'$.
– zzuussee
Jul 25 at 15:07
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and $$A^Ntimes N$$?
– Bongo1234
Jul 25 at 15:24
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and $$A^Ntimes N$$?
– Bongo1234
Jul 25 at 15:24
Try to check the conditions on the matrix $P$ by evaluating the product by hand for the entries in $A'$ this should give you a linear system of equations in variables of $P$ which you then can check for solvability.
– zzuussee
Jul 25 at 18:50
Try to check the conditions on the matrix $P$ by evaluating the product by hand for the entries in $A'$ this should give you a linear system of equations in variables of $P$ which you then can check for solvability.
– zzuussee
Jul 25 at 18:50
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
0
down vote
Yes of course, given that A is invertible we have
$$AP=A' iff P=A^-1A'$$
answered Jul 25 at 15:25
gimusi
65k73583
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes of course, given that A is invertible we have
$$AP=A' iff P=A^-1A'$$
answered Jul 25 at 15:25
gimusi
65k73583
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
add a comment |Â
up vote
0
down vote
Yes of course, given that A is invertible we have
$$AP=A' iff P=A^-1A'$$
answered Jul 25 at 15:25
gimusi
65k73583
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes of course, given that A is invertible we have
$$AP=A' iff P=A^-1A'$$
answered Jul 25 at 15:25
gimusi
65k73583
Yes of course, given that A is invertible we have
$$AP=A' iff P=A^-1A'$$
answered Jul 25 at 15:25
gimusi
65k73583
answered Jul 25 at 15:25
gimusi
65k73583
answered Jul 25 at 15:25
gimusi
65k73583
answered Jul 25 at 15:25
gimusi
65k73583
65k73583
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
add a comment |Â
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and &&A^N times N$$
– Bongo1234
Jul 25 at 15:56
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
I don’t understand what exactly you are asking for, could you please explain better?
– gimusi
Jul 25 at 17:41
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
There exist shift matrices and permutation matrices. If such matrix is multiplied to a vector or another matrix it does the same new range to any matrix. Now I want to have such a matrix for my problem above.
– Bongo1234
Jul 26 at 6:17
add a comment |Â
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What are your thoughts on the problem?
– Parcly Taxel
Jul 25 at 14:57
1
How would you solve $AB=C$ for $B$?
– Ross Millikan
Jul 25 at 15:00
For $A$ invertible, i.e. $a_00cdot a_11cdot a_22neq 0$, i.e. for $a_00, a_11, a_22neq 0$, you can solve the equation $AP=A'$ with $A^-1AP=A^-1A'$, i.e. $P=A^-1A'$.
– zzuussee
Jul 25 at 15:07
thanks,that's no porblem to solve the equation. Is there a general form/matrix which, sets the diagonals into rows for arbitary numbers and $$A^Ntimes N$$?
– Bongo1234
Jul 25 at 15:24
Try to check the conditions on the matrix $P$ by evaluating the product by hand for the entries in $A'$ this should give you a linear system of equations in variables of $P$ which you then can check for solvability.
– zzuussee
Jul 25 at 18:50